Verified solution · Exam P

SOA Exam P Sample Question #404 Solution

This Exam P sample reference tests Marginal Distributions. Summing the joint mass over the other coordinate gives marginal probabilities 51/126, 42/126, and 33/126. Their first two moments yield variance 95/147=0.64626, so choice B.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Marginal Distributions question

Setup

Marginalize the joint mass over all three values of the other coordinate.

\Pr(Y=0)=\frac{24+17+10}{126}=\frac{51}{126}

\Pr(Y=1)=\frac{21+14+7}{126}=\frac{42}{126}

\Pr(Y=2)=\frac{18+11+4}{126}=\frac{33}{126}

Model

Compute the first and second raw moments from the marginal distribution.

E[Y]=\frac{42+2(33)}{126}=\frac67

E[Y^2]=\frac{42+4(33)}{126}=\frac{29}{21}

Compute

Center the second moment to obtain the variance.

\operatorname{Var}(Y)=\frac{29}{21}-\left(\frac67\right)^2

\operatorname{Var}(Y)=\frac{95}{147}=0.6462585034\ldots

Answer

The variance rounds to 0.65.

\boxed{0.65\quad\text{(B)}}

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Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AWith the correct mean 6/7, a variance of 0.56 would imply E[Y²]=0.56+(6/7)²=1.2947, below the directly summed value 29/21=1.3810.
CA variance of 0.75 would imply second moment 1.4847, overstating the contribution from the Y=2 marginal mass.
DA variance of 0.80 would imply second moment 1.5347, again incompatible with the exact marginal calculation 29/21.
EThe value 0.87 is the rounded mean E[Y]=6/7=0.8571, not the variance.

Original practice · fully worked

Original variant: occupied servers after two routes

Two messages are routed independently, and each message is equally likely to go to any of three servers. Let Y be the number of servers that receive at least one of the two messages. Calculate Var(Y).

A
1/9
B
2/9
C
1/3
D
2/3
E
5/3

Variant answer in brief

Both messages use the same server with probability 1/3, producing Y=1; otherwise Y=2. Thus E[Y]=5/3 and E[Y²]=3, so Var(Y)=3-25/9=2/9, choice B.

Setup

The count Y equals one when both messages choose the same server and equals two otherwise.

\Pr(Y=1)=3\left(\frac13\right)^2=\frac13

\Pr(Y=2)=\frac23

Model

Calculate the first and second moments from this two-point distribution.

E[Y]=1\left(\frac13\right)+2\left(\frac23\right)=\frac53

E[Y^2]=1^2\left(\frac13\right)+2^2\left(\frac23\right)=3

Compute

Subtract the square of the mean from the second moment.

\operatorname{Var}(Y)=E[Y^2]-E[Y]^2

\operatorname{Var}(Y)=3-\left(\frac53\right)^2=\frac29

Answer

The occupied-server count has variance 2/9.

\boxed{\frac29\quad\text{(B)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #404 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Marginal Distributions question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: occupied servers after two routes

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution